Subnetting

dimerusticΔίκτυα και Επικοινωνίες

23 Οκτ 2013 (πριν από 3 χρόνια και 9 μήνες)

75 εμφανίσεις

g. babic1
•Given a network IP address, there are three types of problems
involving subnetting:
—Subnettingwhen given a required number of networks
—Subnettingwhen given a required number of clients
—Given an IP address & Subnet Mask, finding original
network range (reverse engineering a subnet problem)
•The following several slides are based on:
https://learningnetwork.cisco.com/servlet/JiveServlet/previewBody/6014-102-1-19236/Subnetting%20Examples.pdf
Subnetting
g. babic2
Subnetting Given Number of Networks (1)
A company would like to break its Class B network IP
address 172.16.0.0 into 60 different subnets. Find
ranges of IP addresses for each subnet and new mask.
oClass B network has 16 host bits Class B subnet mask =
255.255.0.0 = 11111111.11111111.00000000.00000000
o60 = 00111100 we need at least 6 additional network bits
oNew Mask
11111111.11111111.11111(1)00.00000000=255.255.252.0
and bit with parenthesis is the increment bit
oStart with the given network IP address and add the
increment to the subnetted octet:
172.16.0.0
172.16.4.0
172.16.8.0 … etc.
g. babic3
oNow add each end range, which is the last possible IP address
before the next range :
Subnet 1 172.16.0.0 –172.16.3.255
Subnet 2 172.16.4.0 –172.16.7.255
Subnet 3 172.16.8.0 –172.16.11.255
Subnet 4 172.16.12.0 –172.16.15.255


Subnet 60 172.16.236.0 –172.16.239.255
oAssign these ranges to the new networks, but the first and
last address from each range (network / broadcast IP) are
unusable.
Subnetting Given Number of Networks (2)
g. babic4
A company would like to break its Class B private IP
address 172.16.0.0 into as many subnets as possible
provided that they can get
at least
300 clients per
subnet. Find ranges of IP addresses for each subnet
and new mask.
oClass B mask = 11111111.11111111.00000000.00000000
o300 = 100101100 we need at least 9 host bits to remain
oNew Mask
11111111.11111111.111111(1)0.00000000=255.255.254.0
and bit with parenthesis is the increment bit
oStart with the given network address and add the increment to
the subnettedoctet:172.16.0.0
172.16.2.0
172.16.4.0 … etc.
Subnetting Given Number of Hosts (1)
g. babic5
oNow add each end range, which is the last possible IP address
before the next range :
Subnet 1 172.16.0.0 –172.16.1.255
Subnet 2 172.16.2.0 –172.16.3.255
Subnet 3 172.16.4.0 –172.16.5.255
Subnet 4 172.16.6.0 –172.16.7.255


Subnet 128 172.16.254.0 –172.16.255.255
oAssign these ranges to the new networks, but the first and
last address from each range (network / broadcast IP) are
unusable.
Subnetting Given Number of Hosts (2)